• hdu-5683 zxa and xor (位运算)


    题目链接:

    zxa and xor

    Time Limit: 16000/8000 MS (Java/Others)   

     Memory Limit: 65536/65536 K (Java/Others)


    Problem Description
     
    zxa had a great interest in exclusive disjunction(i.e. XOR) recently, therefore he took out a non-negative integer sequence a1,a2,,an of length n.

    zxa thought only doing this was too boring, hence a function funct(x,y) defined by him, in which ax would be changed into y irrevocably and then compute 1i<jn(ai+aj) as return value.

    zxa is interested to know, assuming that he called such function m times for this sequence, then what is the return value for each calling, can you help him?

    tips:1i<jn(ai+aj) means that (a1+a2)(a1+a3)(a1+an)(a2+a3)(a2+a4)(a2+an)(an1+an).
     
    Input
     
    The first line contains an positive integer T, represents there are T test cases.

    For each test case:

    The first line contains two positive integers n and m.

    The second line contains n non-negative integers, represent a1,a2,,an.

    The next m lines, the i-th line contains two non-negative integers x and y, represent the i-th called function is funct(x,y).

    There is a blank between each integer with no other extra space in one line.

    1T1000,2n210^4,1m210^4,0ai,y10^9,1xn,1n,m10^5
     
    Output
     
    For each test case, output in m lines, the i-th line a positive integer, repersents the return value for the i-th called function.
     
    Sample Input
     
    1
    3 3
    1 2 3
    1 4
    2 5
    3 6
     
    Sample Output
     
    4
    6
    8
     
    题意:
     
    给一个数列,每次把a[x]变成y,问每次的fun是多少;
     
    思路:
     
    先算一次fun,再在每次变换的时候运用a^a=0的性质把a[x]+a[i]再异或一边,然后把y+a[i]异或和上去就好了;
     
    AC代码:
     
    //#include <bits/stdc++.h>
    #include <iostream>
    #include <queue>
    #include <cmath>
    #include <map>
    #include <cstring>
    #include <algorithm>
    #include <cstdio>
    using namespace std;
    #define Riep(n) for(int i=1;i<=n;i++)
    #define Riop(n) for(int i=0;i<n;i++)
    #define Rjep(n) for(int j=1;j<=n;j++)
    #define Rjop(n) for(int j=0;j<n;j++)
    #define mst(ss,b) memset(ss,b,sizeof(ss));
    typedef long long LL;
    const LL mod=1e9+7;
    const double PI=acos(-1.0);
    const int inf=0x3f3f3f3f;
    const int N=2e4+25;
    int n,m;
    int a[N],ans;
    int get_ans()
    {
         ans=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                int sum=a[i]+a[j];
                ans^=sum;
            }
        }
    }
    int fun(int fx,int fy)
    {
        for(int i=1;i<=n;i++)
        {
           if(i!=fx) ans^=(a[fx]+a[i])^(fy+a[i]);
        }
        a[fx]=fy;
    }
    int main()
    {
       int t;
       scanf("%d",&t);
       while(t--)
       {
           scanf("%d%d",&n,&m);
           for(int i=1;i<=n;i++)
           {
               scanf("%d",&a[i]);
           }
           int x,y;
           get_ans();
    
           while(m--)
           {
               scanf("%d%d",&x,&y);
               fun(x,y);
               printf("%d
    ",ans);
           }
       }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5499450.html
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